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Prove by epsilon-delta limit definition that: $\lim_{x\to a}e^x=e^a$

My definition of exponential function is $e^x=\lim_{n\to \infty}(1+x/n)^n$. My teacher said that we need to use it but when I use the epsilon-delta definition I have that $$|\lim_{n\to \infty}(1+x/n)^n-\lim_{n\to \infty}(1+a/n)^n|<\epsilon$$ then I don't know how to proceed from here

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    $\begingroup$ What is the definition of $e^x$ you are using in your course? $\endgroup$ – Simon S Nov 6 '14 at 3:41
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    $\begingroup$ what you learned so far? MVT? $\endgroup$ – spatially Nov 6 '14 at 3:42
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    $\begingroup$ The standard way of proving this, I believe, is to simply define $$\log x = \int_1^x \frac{1}{t}\, dt,$$ and then to define $\exp$ as the inverse function to $\log$. By the First Fundamental Theorem of Calculus, $\log$ is continuous, and then it from the theorem that the inverse of a continuous a function is continuous, we know that $\exp$ is continuous. Doing an epsilon-delta proof, however, will depend on the definition of $e^x$ you want to use. $\endgroup$ – user71641 Nov 6 '14 at 3:42
  • $\begingroup$ My definition is $e^x=\lim_{n\to \infty}(1+x/n)^n$. My teacher said that we need to use it but when I use the epsilon-delta definition I have that $|\lim_{n\to \infty}(1+x/n)^n-\lim_{n\to \infty}(1+a/n)^n|<\epsilon$ then I don´t know how to proceed from here $\endgroup$ – user128422 Nov 6 '14 at 3:49
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$|e^x - e^a| = e^a|e^{x - a} - 1|$, then it is enough to show that $\lim_{x \to 0}e^x = 1$.

Use that $$a^n - b^n = (a - b)(a^{n - 1} + ba^{n - 2} + \dots + b^{n - 1})$$ to rewrite the limit as follows: $$\lim_{x \to 0}\big|\lim_{n\to \infty}(1 + xn^{-1})^n - 1\big| = \lim_{x \to 0}\big|\lim_{n \to \infty}(1 + xn^{-1} - 1)((1 + xn^{-1})^{n - 1} + \dots)\big|$$ Now, without loss of generality, assume that $x \le 1$. Noticing that in the second expression between parenthesis we have $n$ terms, all of which are smaller than $e$, we get: $$\lim_{x \to 0}\big|\lim_{n \to \infty}(1 + xn^{-1} - 1)((1 + xn^{-1})^{n - 1} + \dots)\big| \le \lim_{x \to 0}\lim_{n \to \infty}\frac{x}{n}\cdot ne = \lim_{x \to 0}xe = 0$$

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  • $\begingroup$ This assumes that $\lim_{n \to \infty}(1+x/n)^n = e^x$. $\endgroup$ – marty cohen Nov 6 '14 at 4:09
  • $\begingroup$ which is exactly the definition the OP claims to have in the comments under the question... what can I say, sometimes all you've got to do is downvote the question and have fun answering it anyway... :) I like yours btw $\endgroup$ – user67133 Nov 6 '14 at 4:13
  • $\begingroup$ Thanks a lot!! I really appreciate it. I couldn´t find a way. $\endgroup$ – user128422 Nov 6 '14 at 4:15
  • $\begingroup$ You're welcome! but next time please, try to write down all the details we need in order to answer your question! :D As you can see, it generated quite a lot of confusion since no one knew what you were looking for, eheh :) $\endgroup$ – user67133 Nov 6 '14 at 4:16
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    $\begingroup$ Yes! I apologize for the confusion :) Thanks again for your time :D $\endgroup$ – user128422 Nov 6 '14 at 4:22
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One of my math mantras is "always expand around zero".

In this case, that means we want to show that $\lim_{x \to 0} e^{x+a} = e^a $.

But $e^{x+a}-e^a =e^a(e^x-1) =e^a(e^x-e^0) $.

Suppose we can show that $e^x$ is continuous at $0$. This means that, for any $\delta > 0$ we can find an $\epsilon(\delta) > 0$ such that $|x| < \epsilon(\delta)$ implies that $|e^x-1| < \delta$. I'll leave this as an exercise.

To then show that $e^x$ is continuous at $a$, we need to show that, for any $\delta > 0$ we can find an $\epsilon > 0$ such that $|x| < \epsilon$ implies that $|e^{x+a}-e^a| < \delta$. But this is the same as $|e^{x}-1| < \delta e^{-a}$.

Under the assumption that $e^x$ is continuous at zero, consider $\epsilon(\delta e^{-a})$. By the definition of $\epsilon(\delta)$, if $|x| < \epsilon(\delta e^{-a})$ then $|e^x-1| < \delta e^{-a} $, which is the same as $|e^{x+a}-e^x| < \delta $.

Therefore, $e^x$ continuous at zero implies that $e^x$ is continuous everywhere.

A hint for showing $e^x$ is continuous at zero: $e^x-1 =\int_0^x e^t\, dt $.

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  • $\begingroup$ Thanks for your answer @marycohen. I really appreciate your time :) It is very clear $\endgroup$ – user128422 Nov 6 '14 at 4:14

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